Let $k$ be a field, $R$ is a commutative algebra over $k$ and $A$ is an associative algebra over $R$. There is a morphism of commutative algebras $R \to T$. Is it possible to reduce calculation of Hochschild homology $HH_*(A\otimes_R T)$ (over basic field $k$) to $HH_*(A)$? I'm mostly interested in a situation when $T$ is separable over $R$, or even more specific $T=R/I$.

If A is flat over R, or more generally $Tor^R(A,T) = 0$ in positive degrees, then there is a spectral sequence starting with $Tor^{HH_* R}(HH_* A, HH_* T)$ converging to your desired Hochschild term. Does that work in your situation?
–
Tyler LawsonDec 29 '12 at 0:28

Yes, it looks like something that could be very useful for me.
–
Sasha PavlovDec 29 '12 at 0:57

@Tyler Could you please give a reference for this spectral sequence?
–
Sasha PavlovDec 29 '12 at 12:24

1

@Sasha: I'm afraid that I don't have a ready reference. You can prove it by constructing a double complex whose rows are the Hochschild complexes of $A \otimes R^{\otimes q} \otimes T$, with vertical differentials coming from the bar construction. (It's essentially the Hochschild complex of the DGA $A \otimes^{\mathbb L}_R T$.)
–
Tyler LawsonDec 29 '12 at 16:28

1 Answer
1

—theorem (0.1)— you need that $T$ is étale over $R$. This holds in your case only if the ideal $I$ is generated by an idempotent. Notice that "étale" = "unramified + flat". And the notion of unramifiedness is related to the condition of separable ring extension, but it depends on which of the several used definitions of "separable" you consider.

For me, separable means that $T$ is projective over $T \otimes_R T$. I'm aware of this result, but situation when $I$ is generated by an idempotent is not useful for me.
–
Sasha PavlovDec 28 '12 at 23:49

2

Without flatness I have the intuition the result is false. Sorry about that.
–
Leo AlonsoDec 28 '12 at 23:58

Probably won't help, but just a remark in case it's helpful that the etale descent statement extends to smooth descent.
–
David Ben-ZviDec 29 '12 at 0:07